Assuming that you start with 2.00 g of sodium benzoate and 5 mL of 3 M...

Assuming that you start with 2.00 g of sodium benzoate and 5 mL of 3 M HCl, how many molar equivalents of HCl are you using? Use two decimal places.

Expert Solution

mass of sodium benzoate = 2.00 g

molar mass of sodium benzoate= 144.11 g/mol

moles = mass / molar mass

moles of sodium benzoate (C6H5COONa) = 2 / 144.11

= 0.0139

moles of HCl = molarity x volume / 1000

= 3 x 5 / 1000

= 0.015

C6H5COONa + HCl ----------------------> C6H5COOH + NaCl

1 mol 1 mol

0.0139 mol 0.015 mol

here limiting reagent is C6H5COONa

so 0.0139 mol of HCl is used.

molar equivalents of HCl = 0.01 (two decimal places)

queen_honey_blossom answered 2 years ago

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